Question

One particle has a mass of 3.50 x 10-3 kg and a charge of +7.39 μC....

One particle has a mass of 3.50 x 10-3 kg and a charge of +7.39 μC. A second particle has a mass of 7.65 x 10-3 kg and the same charge. The two particles are initially held in place and then released. The particles fly apart, and when the separation between them is 0.184 m, the speed of the 3.50 x 10-3 kg-particle is 180 m/s. Find the initial separation between the particles.

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Answer #1

By Conservation of momentum

m_{1}V_{1A}+m_{2}V_{2A}=m_{1}V_{1B}+m_{1}V_{2B}

Here V1A=V2A=0

0=(3.5*10-3)*180 +(7.65*10-3)*V2B

V2B=-82.353 m/s

By Conservation of energy

12Kqiq2 772 20%, + TA TB

12Kqiq2

(9 x 109)(7.39 x 10-6)2 TA(9 x 109)(7.39 0.184 10-6)2 . Į +x (3.5 x 10-3) x (Эк 109)(7.39 10-6)2 TA 1802 × (7.65 × 10-3) × ×(-82.353)2

r = 5.76 × 10-3m

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